A model train, with a mass of $8 kg$ , is moving on a circular track with a radius of $1 m$ . If the train's rate of revolution changes from $5/8 Hz$ to $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?

Question

1694 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T20:50:42

$123.37N$

The centripetal force of an object is given by

$F = (mv^2)/r$

You can also find $v$ by

$v = 2pirf$

where $f$ is the frequency of revolution.

Substitute this into the above equation,

$F = (m(2pirf)^2)/r = (4mpi^2r^2f^2)/r$

$=4mrpi^2f^2$

or, since we're looking at the change in centripetal force from the change in frequency, then

$DeltaF = 4mrpi^2(Deltaf)^2$

where $Deltaf$ is $5/4Hz - 5/8Hz = 5/8Hz$

Therefore, using the values that we know,

$DeltaF = 4xx8xx1xxpi^2xx(5/8)^2$

$= 123.37N$

A model train, with a mass of 8 kg8 kg, is moving on a circular track with a radius of 1 m1 m. If the train's rate of revolution changes from 5/8 Hz5/8 Hz to 5/4 Hz5/4 Hz, by how much will the centripetal force applied by the tracks change by?